Number Sequence(找规律，，，，，反正我没找到，递归超内存了....)

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 178534    Accepted Submission(s): 44354

Problem Description

A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

Sample Input

1 1 31 2 100 0 0

Sample Output

25

一看这题立马写了个递归然后出了个这个Memory Limit Exceeded，，，内存超了，，

#include <stdio.h>#include <iostream>#include <algorithm>#include <stdlib.h>#include <string.h>using namespace std;int f(int A, int B, int n){    if(n==1) return 1;    if(n==2) return 1;    return (A*f(A, B, n-1)+B*f(A, B, n-2))%7;}int main(){    int n, A, B;    while(scanf("%d%d%d",&A,&B,&n), A&&B&&n){        printf("%d\n",f(A,B,n));    }    return 0;}

然后发现应该是个循环，再写被T了；

#include <stdio.h>#include <iostream>#include <algorithm>#include <stdlib.h>#include <string.h>using namespace std;int main(){    int n, A, B;    while(scanf("%d%d%d",&A,&B,&n)){        if(A==0 && B==0 && n==0) break;        int ff[1000];        ff[1]=ff[2]=1;        int i=3;        for(i=3; i<=n; i++){            ff[i]=(A*ff[i-1]+B*ff[i-2])%7;           // printf("%d ",ff[i]);            if(ff[i]==1&&ff[i-1]==1) break;        }        //printf("\n");        if(i>n) printf("%d\n", ff[n]);        else{        int cnt=i-2;        int ans;        if(n%cnt==0) ans=cnt;        else ans=n%cnt;        printf("%d\n",ff[ans]);        }    }    return 0;}

然后再改，RE，，，，

#include <stdio.h>#include <iostream>#include <algorithm>#include <stdlib.h>#include <string.h>using namespace std;int main(){    int n, A, B;    while(scanf("%d%d%d",&A,&B,&n)){        if(A==0 && B==0 && n==0) break;        int ff[100];        ff[1]=ff[2]=1;        int i=3;        for(i=3; i<=n; i++){            ff[i]=(A*ff[i-1]+B*ff[i-2])%7;            //printf("%d ",ff[i]);            if(ff[i]==1&&ff[i-1]==1) break;        }       // printf("\n");        if(i>n) printf("%d\n", ff[n]);        else{        int cnt=i-2;       // printf("cnt: %d\n",cnt);        ff[0]=ff[cnt];        printf("%d\n",ff[n%cnt]);        }    }    return 0;}

当我把for循环里的n改成50就A了，改成100还是RE，醉了；

#include <stdio.h>#include <iostream>#include <algorithm>#include <stdlib.h>#include <string.h>using namespace std;int main(){    int n, A, B;    while(scanf("%d%d%d",&A,&B,&n)){        if(A==0 && B==0 && n==0) break;        int ff[100];        ff[1]=ff[2]=1;        int i=3;        for(i=3; i<=50; i++){        //就是这里的50，改成n就RE，，，，，，，            ff[i]=(A*ff[i-1]+B*ff[i-2])%7;            //printf("%d ",ff[i]);            if(ff[i]==1&&ff[i-1]==1) break;        }       // printf("\n");        if(i>n) printf("%d\n", ff[n]);        else{        int cnt=i-2;       // printf("cnt: %d\n",cnt);        ff[0]=ff[cnt];        printf("%d\n",ff[n%cnt]);        }    }    return 0;}

最后附上大神的AC代码，就是强啊，咋想到的呢，，，，，

#include <iostream>using namespace std;int main(){    int A,B,n,i;    int a[48];    cin >> A >> B >> n;    while( A!=0 || B!=0 || n!=0)    {        a[0]=a[1]=1;        for(i=2;i<48;i++)            a[i] = ( A * a[i-1] + B * a[i-2]) % 7;        cout << a[(n-1)%48] <<endl;//48个数必定循环一遍，或以48的因子数一个循环......膜了；         cin >> A >> B >> n;    }    return 0;}